poj 1466 Girls and Boys

Girls and Boys

Time Limit: 5000MS		Memory Limit: 10000K
Total Submissions: 12857		Accepted: 5723

Description

In the second year of the university somebody started a study on the romantic relations between the students. The relation "romantically involved" is defined between one girl and one boy. For the study reasons it is necessary to find out the maximum set satisfying the condition: there are no two students in the set who have been "romantically involved". The result of the program is the number of students in such a set.

Input

The input contains several data sets in text format. Each data set represents one set of subjects of the study, with the following description: 

the number of students 
the description of each student, in the following format 
student_identifier:(number_of_romantic_relations) student_identifier1 student_identifier2 student_identifier3 ... 
or 
student_identifier:(0) 

The student_identifier is an integer number between 0 and n-1 (n <=500 ), for n subjects.

Output

For each given data set, the program should write to standard output a line containing the result.

Sample Input

7
0: (3) 4 5 6
1: (2) 4 6
2: (0)
3: (0)
4: (2) 0 1
5: (1) 0
6: (2) 0 1
3
0: (2) 1 2
1: (1) 0
2: (1) 0

Sample Output

5
2

翻译：给出n个人，这些人相互之间存在爱情，现在要从这n个人中挑出一个集合，使得集合中的人相互之间都没有爱慕关系，强行让他们都变成单身狗，问这个集合最大的人数。
思路：题意就是最大独立集问题，且最大独立集=V-最大匹配，V是所有顶点的个数，此题中即为n.
AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<queue>
#include<set>
#include<vector>
#include<cstring>
#include<string>
using namespace std;
#define INF 0x3f3f3f3f
const int N_MAX =500;
int V;//点的个数
vector<int>G[N_MAX];
int match[N_MAX];
bool used[N_MAX];
void add_edge(int u, int v) {
    G[u].push_back(v);
    //G[v].push_back(u);
}

bool dfs(int v) {
    used[v] = true;
    for (int i = 0; i < G[v].size(); i++) {
        int u = G[v][i], w = match[u];
        if (w < 0 || !used[w] && dfs(w)) {
            match[v] = u;
            match[u] = v;
            return true;
        }
    }
    return false;
}

int bipartite_matching() {
    int res = 0;
    memset(match, -1, sizeof(match));
    for (int v = 0; v < V; v++) {
        if (match[v] < 0) {
            memset(used, 0, sizeof(used));
            if (dfs(v))
                res++;
        }
    }
    return res;
}


int main() {
    int n;
    while (scanf("%d",&n)!=EOF) {
        V = n;
        for (int i = 0; i < n;i++) {
            int x, n1;
            scanf("%d: (%d) ",&x,&n1);
            if (n1)for (int j = 0; j < n1; j++) {
                int y;
                scanf("%d",&y);
                add_edge(x,y);
            }
        }
        printf("%d
",V-bipartite_matching());
        for (int i = 0; i < V;i++) {
            G[i].clear();
        }
    }
    return 0;
}